of 1 71
The Planets, AI Elements, And Life!
By Ian Beardsley!
Copyright © by Ian Beardsley 2020"
of 2 71
"
of 3 71
Table of Contents
Introduction………………………………………4!
Silicon and Carbon………………………………4!
The Planets and the AI Elements………………6!
Germanium and Carbon………………………..17!
The Fundamental AI BioEquations…………….20!
The Generalized Equation………………………22!
The Protoplanetary Disc…………………………24!
Mercury And Bone……………………………….26!
Masculine and Feminine in AI and Life………..33!
The Means of Si and Ge………………………..34!
The Pattern……………………………………….45!
Modeling The Solar System……………………49!
Hydrogen…………………………………………54!
The Scale of Things……………………………..59!
The Nature Of Our Equation……………………63!
The Orbits of the Planets……………………….68!
Proposal For Mars Mission Logo………………70!
of 4 71
Introduction
In studying the relationship between biological life and artificial intelligence (AI) I have met with
some very interesting equations in terms of molar masses, densities, and atomic radii of the
characteristic elements and their compounds that are characteristic of them. I was working on
this and doing a computation that errantly produced an equation I found interesting
nonetheless, and I recognized it to be exactly the Mars average orbit in astronomical units. I
then pursued the other terrestrial planets Venus and Earth itself. I wanted their to be equation
for them in terms of silicon and germanium as well (core AI elements) and to be of the form of
the Mars equation with polynomials, as I was thinking perhaps the orbits of the planets are
related to polynomial expressions of the AI elements. And indeed they were to two places after
the decimal in the case of Venus and nearly two places after the decimal in the case of Earth. !
I do not think it is silly, but perhaps these polynomials predicting the orbits of the terrestrial
(inner solid planets) in terms of the elements silicon and germanium, can be derived by
considering the protoplanetary disc from which the planets formed, to have had a molar mass
as a mixture around the mean between silicon and germanium. !
Indeed this is how my project began in studying the connection between biological life and AI
which I found was in the golden ratio conjugate. As follows:!
Silicon and Carbon
The golden ratio and the golden ratio conjugate are the solution of the quadratic!
that meets the conditions and a=b+c!
Where and , .!
We guess that artificial intelligence (AI) has the golden ratio, or its conjugate in its means
geometric, harmonic, and arithmetic by molar mass by taking these means between doping
agents phosphorus (P) and boron (B) divided by semiconductor material silicon (Si) :!
!
!
!
Which can be written!
(
a
b
)
2
−
a
b
− 1 = 0
a
b
=
b
c
Φ =
a
b
ϕ =
5 − 1
2
ϕ =
1
Φ
2PB
P + B
1
Si
=
2(30.97)(10.81)
30.97 + 10.81
1
28.09
= 0.57
0.65 + 0.57
2
= 0.61 ≈ ϕ
of 5 71
!
We see that the biological elements, H, N, C, O compared to the AI elements P, B, Si is the
golden ratio conjugate (phi) as well:!
!
So we can now establish the connection between artificial intelligence and biological life:!
!
Which can be written:!
!
Where HNCO is isocyanic acid, the most basic organic compound. We write in the arithmetic
mean:!
!
Which is nice because we can write in the second first generation semiconductor as well
(germanium) and the doping agents gallium (Ga) and arsenic (As):!
!
Where!
!
Where ZnSe is zinc selenide, an intrinsic semiconductor used in AI, meaning it doesn’t require
doping agents. We now have:!
!
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
C + N + O + H
P + B + Si
≈ ϕ
(P + B + Si)
PB(P + B) + 2PB
2(P + B)Si
≈ (C + N + O + H )
PB
[
P
Si
+
B
Si
+ 1
]
+
2PB
P + B
[
P
Si
+
B
Si
+ 1
]
≈ 2HCNO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ 3HNCO
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
Zn
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
PB
(
Zn
Se
)
+
2PB
P + B
(
Zn
Se
)
+
P + B
2
(
Zn
Se
)
≈ HNCO
of 6 71
The Planets And AI Elements
It would seem if we construct an AI periodic table that uses the primary semiconductor
materials Ge, and Si, and their doping agents boron, gallium, phosphorus and arsenic, as such
pulling them out of the periodic table…!
!
We see by dividing the molar mass of Si semiconductor, and carbon primary to biological life
we get 28.09/12.01=2.33 and for p-type semiconductors Ga/B=6.449583 and if we divide the
Ga by aluminum we get 69.72/26.98=2.584. Or, if we divide n-type by n-type (As/P) we get
74.92/30.97=2.419. We have trouble getting 3/2 in all of these cases. Let us try the finest
conductors for making electrical wire gold and silver, we get 1.8, or silver divided by most
widely used electric wire copper which is 1.6974. Let us move out of our AI table again and try
the intrinsic semi-conductor zinc selenide. We get Se/Zn=78.96/65.38=1.2. Let us go
horizontal, P/Al=1.147, or, As/Ga=74.92/69.72=1.07. Let us go horizontal as well biology’s
oxygen and carbon, we get 1.16. We keep dancing around 3/2. However, if we leave the
microscopic world and go to the macroscopic, the planets, we have the the 3/2 appears in the
ratio of the terrestrial planets Earth, and Mars, which is 1.52 which is:!
!
This equation, which I discovered by making an error in a computation, I saw as interesting
because it is reminiscent of the harmonic mean:!
!
1.52 =
2SiGe
Ge
2
− Si
2
H =
2ab
a + b
of 7 71
However, a and b are squared in the denominator and their difference is taken instead of their
sum, reminiscent of statistical analysis (the method of least squares) used to find the deviation
of b from a. Plotting this using WolframAlpha reveals an interesting form indeed; Something like
a propeller:!
!
Compare this to the plot of the actual harmonic mean:!
of 8 71
We have Mars is:!
!
!
We proceed to look at what the equation for the earth might look like (1.00 astronomical units).
It is is a related polynomial that followed in my mind from the above. I simply wrote the
denominator with the square of the differences instead of the the difference of the squares, and
did not multiply the product between Si and Ge by 2. It works nearly exact as well:!
!
Which can be written!
and has plot!
1.52 =
2SiGe
Ge
2
− Si
2
1.01 =
SiGe
(Ge − Si )
2
SiGe
(Ge
2
− 2SiGe + Si
2
)
of 9 71
We look at this and guess Venus is:!
!
Or,!
!
It turns out it is the arithmetic mean between the two which is:!
!
And has plot!
All of this using Si=28.09 g/mol, and Ge=72.61g/mol"
2SiGe
Ge
2
+ Si
2
2SiGe
Ge
2
=
2Si
Ge
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
of 10 71
Mars !
Earth !
!
1.52 =
2SiGe
Ge
2
− Si
2
1.01 =
SiGe
(Ge − Si )
2
of 11 71
Venus !
!
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
of 12 71
Earth equation"
of 13 71
I said the Venus orbit was !
!
And, that I arrived at it by averaging my two estimates, which were!
and,…!
!
It is then simple to attain the equation for mercury because its average distance from the sun is
0.4 astronomical units, but since it has a highly eccentric orbit it can drop to below this or go
above it. The orbit can be described then by one half either one of these equations. Their plots
are:!
!
"
0.72 =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
2
2SiGe
Ge
2
+ Si
2
2SiGe
Ge
2
=
2Si
Ge
of 14 71
!
!
SiGe
Ge
2
+ Si
2
= 0.3
SiGe
Ge
2
=
Si
Ge
= 0.387
of 15 71
Now we move to the outer planets, most of which are gas giants. It becomes obvious what we
do for Jupiter, which is at 5.2 AU from the sun on the average. We take the earth equation:!
!
And, invert it, put the polynomial in the numerator, and the product in the denominator, but to
follow previous patterns, reverse the negative sign and we have:!
!
It is incredible how close it is to 5.2 AU, and it can be made to work perfectly by adjusting the
coefficients of the polynomial, however we want as above because we want the polynomial in
the numerator to be a perfect square. That is, it can be factored!
!
Its plot is!
!
SiGe
(Ge
2
− 2SiGe + Si
2
)
= 1.01
Si
2
+ 2SiGe + Ge
2
SiGe
= 4.97AU ≈ 5AU
(Si + Ge)
2
SiGe
= 4.97
of 16 71
We now move out to Saturn and the solution is simple. Just as the Mercury equations were half
the Venus equations, Saturn (9 AU) is twice the Jupiter equation:!
!
Which of course, like the Jupiter equation can have its coefficients adjusted. It plot is:!
!
The next planet out, Uranus (19.2 AU), is clearly about twice the Saturn Equation:!
!
And, Neptune (30.33 AU) is clearly three times the Saturn Equation:!
!
Pluto has been declassified as a planet. !
Can I write one equation that becomes all of these equations in terms of planetary orbital
number? I seem to understand my reasoning in arriving at these, so I think I can write an
algorithm for computer code that produces them as such.!
2(Si + Ge)
2
SiGe
= 9.94
4(Si + Ge)
2
SiGe
= 19.88
6(Si + Ge)
2
SiGe
= 29.82
of 17 71
Germanium And Carbon
But just as in the outset of this paper when I presented the connection of AI to biological life
and showed my results for silicon doped with phosphorus and boron, I have similar results for
germanium doped gallium and arsenic, silicon doped with gallium and arsenic, and so on.
Therefore, since the planet equations are all in silicon and germanium, I can bring in the doping
agents phosphorus, boron, gallium, arsenic,…and write out the planetary equations in terms of
them, and, in terms of the biological. That would be an immense task and would better be
done on computers with sophisticated math software. I further wrote…!
We could begin with semiconductor germanium (Ge) and doping agents gallium (Ga) and
Phosphorus (P) and we get a similar equation:!
, !
In grams per mole. Then we compare these molar masses to the molar masses of the
semiconductor material Ge:!
!
!
Then, take the arithmetic mean between these:!
!
We then notice this is about the golden ratio conjugate, , which is the inverse of the golden
ratio, . . Thus, "
2Ga P
Ga + P
= 42.866
Ga P = 46.46749
2Ga P
Ga + P
1
Ge
=
42.866
72.61
= 0.59
Ga P
1
Ge
=
46.46749
72.61
= 0.64
0.59 + 0.64
2
= 0.615
ϕ
Φ
ϕ ≈
1
Φ
of 18 71
we have!
1. !
2. !
This is considering the elements of artificial intelligence (AI) Ga, P, Ge, Si. Since we want to find
the connection of artificial intelligence to biological life, we compare these to the biological
elements most abundant by mass carbon (C), hydrogen (H), nitrogen (N), oxygen (O),
phosphorus (P), sulfur (S). We write these CHNOPS (C+H+N+O+P+S) and find:!
!
A similar thing can be done with germanium, Ge, and gallium, Ga, and arsenic, As, this time
using CHNOPS the most abundant biological elements by mass:!
!
!
!
!
!
We can also make a construct for silicon doped with gallium and phosphorus:!
!
!
Ga P(Ga + P) + 2G a P
2(Ga + P)Ge
≈ ϕ
Ga P(Ga + P) + 2G a P
2(Ga + P)Si
≈ Φ
CHNOPS
Ga + As + Ge
≈
1
2
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
≈ CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
Ga As
(
O
S
)
+
2Ga As
Ga + As
(
O
S
)
+
Ga + As
2
(
O
S
)
≈ CHNOPS
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
Ga As(Ga + As) + 2Ga As
2(Ga + As)Ge
≈ 1
C + H + N + O + P + S
Ga + As + Ge
≈
1
2
(C + N + O + H ) ≈
2(Ga + P)Si
Ga P(Ga + P) + 2G a P
(P + B + Si)
HNCO ≈
2(Ga + P)Si
(Ga + P)
[
Ga P +
2GaP
Ga + P
]
(P + B + Si)
of 19 71
!
And for germanium doped with gallium and phosphorus:!
!
!
"
HNCO ≈
2(P + B + Si)Si
Ga P +
2GaP
Ga + P
Ga P(Ga + P) + 2G a P
2(Ga + P)Ge
≈ ϕ
[
Ga P +
2Ga P
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
Ga P
(
B
S
)
+
2Ga P
Ga + P
(
B
S
)
+
Ga + P
2
(
B
S
)
≈ HNCO
of 20 71
The Fundamental AI BioEquations
!
!
!
!
!
!
!
!
!
!
!
"
[
PB +
2PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
≈ CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
[
Ga P +
2Ga P
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
HNCO ≈
2(P + B + Si)Si
Ga P +
2GaP
Ga + P
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
Ga As(Ga + As) + 2Ga As
2(Ga + As)Ge
≈ 1
Ga P(Ga + P) + 2G a P
2(Ga + P)Ge
≈ ϕ
Ga P(Ga + P) + 2G a P
2(Ga + P)Si
≈ Φ
C + N + O + H
P + B + Si
≈ ϕ
C + H + N + O + P + S
Ga + As + Ge
≈
1
2
Zn
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
of 21 71
"
of 22 71
The Generalized Equation
We want to write a generalized equation for the expressions of the planets. We make our job
simpler by breaking it up into two equations; one for the inner terrestrial planets, one for the
outer planets. We see the equation for the outer planets is of inverse form to that of the inner
planets, the quadratics in the inner planets are in the denominator, and, in the numerator for the
outer planets. The inner and outer planets are separated by the asteroid belt. For the inner
planets, 3 Earth, 4 Mars, we have!
!
We see the equation is P_n (n=number of planet) is equal to a ratio between the product of Si
and Ge, divided by a quadratic with the middle term multiplied by some integer i,…!
!
Thus, when n=3 then c=1 and i=2 and we have the Earth equation:!
!
And, when n=4 (Mars) we have c=2 and i=0, and we reverse the sign in Si in the denominator.!
!
For the outer planets, 5 Jupiter, 6 Saturn, 7 Uranus, 8 Neptune, we have one simple equation
for all of them:!
!
Thus when n=5 for Jupiter we have i=1 yields its equation:!
!
For n=6 for Saturn we have i=2:!
P
n
=
cSiGe
(Ge
2
− iSiGe + Si
2
)
P
3
=
1SiGe
(Ge
2
− 2SiGe + Si
2
)
P
4
=
2SiGe
(Ge
2
− 0SiGe + Si
2
)
P
n
=
i(Si + Ge)
2
SiGe
P
5
=
(Si + Ge)
2
SiGe
of 23 71
!
For n=7 for Uranus we have i=4 to get!
!
And finally for Neptune planet 8, i=6:!
!
P
6
=
2(Si + Ge)
2
SiGe
P
7
=
4(Si + Ge)
2
SiGe
P
8
=
6(Si + Ge)
2
SiGe
of 24 71
The Protoplanetary Disc
If our protoplanetary disc, as I suggested, can be taken as having a molar mass that is the
mean between germanium and silicon, then it can be taken as having a density that is the
mean between the density of germanium and silicon.!
"
of 25 71
!
Taking the protoplanetary disc as a thin disc we integrate from its center to the edge, with
density decreasing linearly to zero at the edge. Thus, if the density function is given by!
!
And, our integral is!
!
!
!
The mass of the solar system adding up all the planets yields!
!
That accounts for!
82% of the mass of the solar system not including the sun, that is, of the
protoplanetary disc surrounding the sun.!
Using germanium alone, we get,!
!
If we weight the mixture of silicon and germanium as 1/3 and 2/3, then we have!
!
Which is very close.!
93%!
This is all very good, because I only used the planets and asteroids."
Si + Ge
2
=
2.33 + 5.323
2
= 3.8265g /cm
3
ρ(r ) = ρ
0
(
1 −
r
R
)
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ
M =
πρ
0
R
2
3
π(3.8265)(7.4 × 10
14
)
2
3
= 2.194 × 10
30
grams
M = 2.668 × 10
30
grams
2.194
2.668
100 =
π(5.323)(7.4 × 10
14
)
2
3
= 3.05 × 10
30
grams
π(4.32467)(7.4 × 10
14
)
2
3
= 2.48 × 10
30
grams
2.48
2.668
100 =
of 26 71
Weighting silicon and germanium as 1/4 and 3/4 we have!
!
Which accounts for!
98%!
Of the mass of the solar system (very accurate).!
This mixture of 1/4 to 3/4 is a combination that exists in the Earth atmosphere which is
approximately the mixture of oxygen to nitrogen. The earth atmosphere can be considered a
mixture of chiefly O2 and N2 in these proportions:!
Air is about 25% oxygen gas (O2) by volume and 75% nitrogen gas (N2) by volume meaning
the molar mass of air as a mixture is:!
!
By molar mass the ratio of air to H20 (water) is about the golden ratio:!
!
I am not saying the solar system is a thin disk with density of the weighted mean somewhere
between silicon and germanium, but that it can be modeled as such, though if the
protoplanetary disk that eclipses epsilon aurigae every 27 years is any indication of what a
protoplanetary cloud is like, it is a thin disk in the sense that it is about 1 AU thick and 10 AU in
diameter. This around a star orbiting another star.!
Mercury and Bone
We said Mercury orbit was in the ratio of Si to Ge. This has an interesting occurrence I found
earlier in bone in its connection to Si and Ge as well. I wrote!
Density of silicon is Si=2.33 grams per cubic centimeter.!
Density of germanium is Ge=5.323 grams per cubic centimeter.!
Density of hydroxyapatite is HA=3.00 grams per cubic centimeter.!
This is!
where !
π(4.4.57475)(7.4 × 10
14
)
2
3
= 2.623 × 10
30
grams
2.623
2.668
100 =
0.25O
2
+ 0.75N
2
≈ air
air
H
2
O
≈ Φ
3
4
Si +
1
4
Ge ≈ H A
HA = Ca
5
(PO
4
)
3
OH
of 27 71
Where HA is the mineral component of bone, Si is an AI semiconductor material and Ge is an
AI semiconductor material. This means!
!
The harmonic mean between Si and Ge is HA,…!
!
This is the sextic,…!
!
Which has a solution!
!
Where x=Si, and y=Ge. It can be solved with the online Wolfram Alpha computational engine.!
This presents itself as a series of fundamental mathematical operations, thus leading me to
suggest that life and AI can be taken as mathematical in structure:!
Multiplying Binomials!
Completing The Square!
The Quadratic Formula!
Ratios!
Proportions!
The Golden Ratio!
The Square Root of Two!
The Harmonic Mean!
Si
HA
Si +
[
1 −
Si
HA
]
Ge = H A
2SiGe
Si + Ge
≈ H A
x
2
(x + y)
4
− x y(x + y)
4
+ 2x y
2
(x + y)
3
− 4x
2
y
2
(x + y)
2
= 0
Si
Ge
=
1
2 + 1
of 28 71
We begin with
!
!
!
!
!
!
Multiplying Binomials and The Quadratic Formula!
!
!
We see that the square of the binomial is a quadratic where the third term is the square of one
half the middle coefficient. This gives us a method to solve quadratics called completing the
square:!
!
!
!
!
!
Si
HA
Si +
[
1 −
Si
HA
]
Ge = H A
Si
2
HA
+ Ge −
Si
HA
Ge ≈ H A
1
HA
Si
2
−
Ge
HA
Si + Ge ≈ H A
1
HA
2
Si
2
−
Ge
HA
2
Si +
Ge
HA
≈ 1
1
HA
2
Si
2
−
Ge
HA
2
Si +
Ge
HA
− 1 ≈ 0
1
HA
2
Si
2
−
Ge
HA
2
Si +
[
Ge
HA
− 1
]
= 0
(x + a)(x + a) = x
2
+ 2a x + a
2
(x + a)
2
= x
2
+ 2a x + a
2
a x
2
+ bx + c = 0
a x
2
+ bx = − c
x
2
+
b
a
x = −
c
a
(
1
2
b
a
)
2
=
1
4
b
2
a
2
x
2
+
b
a
x +
1
4
b
2
a
2
= −
c
a
+
1
4
b
2
a
2
of 29 71
!
!
!
!
!
!
!
!
!
!
!
!
!
(
x +
1
2
b
a
)
2
=
b
2
− 4ac
4a
2
x +
b
2a
=
±
b
2
− 4ac
2a
x =
−b
±
b
2
− 4ac
2a
1
HA
2
Si
2
−
Ge
HA
2
Si +
[
Ge
HA
− 1
]
= 0
x =
−b
±
b
2
− 4ac
2a
a =
a
HA
2
b = −
Ge
HA
2
c =
[
Ge
HA
− 1
]
b
2
− 4ac =
Ge
2
HA
4
− 4
1
HA
2
[
Ge
HA
− 1
]
=
Ge
2
HA
4
−
4Ge
HA
3
+
4
HA
2
=
1
HA
2
[
Ge
2
HA
2
−
4Ge
HA
+ 4
]
b
2
− 4ac =
1
HA
(
Ge
HA
− 2
)
2
x =
Ge
HA
2
±
1
HA
[
Ge
HA
− 2
]
2
HA
2
=
1
2
Ge
±
1
2
HA
[
Ge
HA
− 2
]
=
1
2
Ge
±
1
2
Ge − H A
of 30 71
!
!
Completing The Square!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
Si =
1
2
Ge +
1
2
Ge − H A
Si = Ge − HA
Si ≈ Ge − H A
HA ≈
2SiGe
Si + Ge
Si ≈ Ge −
2SiGe
Si + Ge
(Si + Ge)Ge
Si + Ge
−
(Si + Ge)Si
Si + Ge
−
2SiGe
Si + Ge
= 0
Ge
2
− 2SiGe − Si
2
Si + Ge
= 0
x
2
− 2x y − y
2
= 0
x
2
− 2x y = y
2
x
2
− 2x y + y
2
= 2y
2
(x − y)
2
= 2y
2
x − y =
±
2y
x = y + 2y
x = y(1 + 2)
x
y
= 1 + 2
y
x
=
1
2 + 1
Si
Ge
≈
1
2 + 1
of 31 71
The Golden Ratio
A ratio is and a proportion is which means a is to b as b is to c.!
The Golden Ratio !
and. !
or !
!
!
!
!
!
!
!
!
a
b
a
b
=
b
c
(
Φ
)
a
b
=
b
c
a = b + c
ac = b
2
c =
b
2
a
a = b +
b
2
a
b
2
a
− a + b = 0
b
2
a
2
− 1 +
b
a
= 0
(
b
a
)
2
+
b
a
− 1 = 0
(
b
a
)
2
+
b
a
+
1
4
= 1 +
1
4
(
b
a
+
1
2
)
2
=
5
4
b
a
= −
1
2
±
5
2
b
a
=
5 − 1
2
a
b
=
5 + 1
2
ϕ =
5 − 1
2
Φ =
5 + 1
2
ϕ =
1
Φ
of 32 71
The mineral component of bone hydroxyapatite (HA) is!
!
The organic component of bone is collagen which is!
!
We have!
!
!
!
!
%!
!
!
Thus,..!
=!
(Mercury Orbit)/(Earth Orbit)!
Ca
5
(PO
4
)
3
OH = 502.32
g
mol
C
57
H
91
N
19
O
16
= 1298.67
g
mol
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= 0.386795722
ϕ = 0.618033989
1 − ϕ = 0.381966011
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
≈ (1 − ϕ)
0.381966011
0.386795722
100 = 98.75
Si
Ge
=
28.09
72.61
= 0.386861314 ≈ (1 − ϕ)
Si
Ge
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
Si
Ge
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
of 33 71
The Masculine and Feminine in AI and Life
I have further found the relationship between the masculine and feminine in terms of AI to be in
inverse but complimentary relationship to one another. This again is in the ratio of Mercury orbit
to Earth orbit. I wrote:
We consider the female sex hormone estradiol (estrogen , E):!
!
And the male sex hormone testosterone (T):!
!
And, cholesterol (Ch) from which both are made:!
!
And notice,…!
!
And we consider the semiconductor materials used to make AI:!
!
And write,…!
!
!
!
We notice that the masculine (T) is in inverse relation to the feminine (E), but that the two add
up to on whole (Ch) in that the masculine has coefficient 1-Si/Ge and the feminine has
coefficient 1-Ge/Si. This expresses the inverse relationships between man and woman."
C
18
H
24
O
2
= 272.38g /m ol
C
19
H
28
O
2
= 288.42g /m ol
C
27
H
46
O = 386.65g/mol
Ch + T
E
= 2.5
Ge
Si
= 2.6
Ch + T
E
=
Ge
Si
T =
Ge
Si
E − Ch
E =
Si
Ge
(T + Ch)
T
(
1 −
Si
Ge
)
+ E
(
1 −
Ge
Si
)
= Ch
(
Si
Ge
− 1
)
of 34 71
The Means of Si and Ge
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al) is
in boron’s place. This results in an interesting equation.
The differential across germanium crossed with silicon plus the differential across silicon crossed
with germanium normalized by the product between silicon and germanium is equal to the boron
divided by the average between the germanium and the silicon. The equation has nearly 100%
accuracy:
"
Si(As − Ga) + Ge(P − Al )
SiG e
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
of 35 71
By Molar Mass!
We found (Beardsley, Mathematical Structure, 2020) that the differential across silicon (P-Al)
times germanium (Ge) over boron (B) plus the differential across germanium (As-Ga) times
silicon (Si) over boron (B) was equal to the harmonic mean between Si and Ge. This was
interesting because aluminum is used as what I called a dummy doping agent element, which
when inserted predicts the actually doping agent boron, that seems out of place in the periodic
table where the core artificial intelligence elements are concerned. This is written:!
Stokes Theorem states:
We know the harmonic mean H of a function is
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
2SiGe
Si + Ge
∫
S
( ∇ ×
⇀
u ) ⋅ d
⇀
S =
∮
C
⇀
u ⋅ d
⇀
r
∇ ×
⇀
u =
⇀
i
⃗
j
⇀
k
∂
∂x
∂
∂y
∂
∂z
u
1
u
2
u
3
⃗
i
⃗
j
⃗
k
∂
∂x
∂
∂y
∂
∂z
0
Si
B
(Ga)z
Si
B
(As)y
=
Si
B
(As − Ga)
⃗
i
⃗
i
⃗
j
⃗
k
∂
∂x
∂
∂y
∂
∂z
Ge
B
(Al )z 0
Ge
B
(P)x
=
Ge
B
(P − Al )
⃗
j
⃗
u = (u
1
, u
2
, u
3
)
⃗
v = (v
1
, v
2
, v
3
)
⃗
u = 0
⃗
i +
Si
B
(Ga)z
⃗
j +
Si
B
(As)y
⃗
k
⃗
u = 0
⃗
i +
Si
B
(Ga)z
⃗
j +
Si
B
(As)y
⃗
k
of 36 71
And, that the arithmetic mean A of a function is
We have
But, we want to use Stokes theorem so we want the integral in the numerator. So, we make the
approximation
And, we have
But, this is only 80% accurate. We find it is very accurate if we say
Which yields
By Density
We ask if the asymmetry in the AI elements in the periodic table due to boron results in a
dynamic equation by molar mass, then does it as well by density? While molar mass is due to the
composition of elements, density is due to the balance between the strong nuclear force holding
protons together balanced by their electric forces that are mutually repulsive.
The density of boron is 2.340 grams per cubic centimeter, that of phosphorus (white
phosphorus) is 1.88 grams per cubic centimeter and gallium is 5.904 grams per cubic centimeter.
Arsenic is 5.7 grams per cubic centimeter, germanium is 5.323 grams per cubic centimeter and
aluminum is 2.7 grams per cubic centimeter. We have the following scenario:
H =
1
1
b − a
∫
b
a
f (x)
−1
d x
A =
1
b − a
∫
b
a
f (x)d x
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
∫
Ge
Si
dx
x
H ≈ A
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
x d x
f (x) =
4
5
x
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
4
5
∫
Ge
Si
x d x
of 37 71
Again we see boron breaks the symmetry in that period three densities are on the same order and
period 4 densities are on the same order, but that of boron is almost the same as silicon in period
three. We see that semiconductor material Si is the the average between doping agent P and
would be doping agent aluminum that takes the place of boron and, that, the average doping
agent Ga and semiconductor material Ge is approximately the average of doping agent As. Thus
we have:
Ga-As=0.204 (differential across Ge)
Al-P=0.82 (dummy differential across Si)
Si~B
Ge/Si~B
(0.82)
Si
B
= 0.816
(0.204)
Ge
B
= 0.464
0.816
0.464
= 1.7586
GeSi = 3.52
3.52
2
= 1.76 ≈ 1.7586
(Al − P)
Ga − As
B ≈
GeSi
2
Ge
of 38 71
And, we have
The first factor takes the form of the harmonic mean between a and b, H:
And the second term takes the form of the geometric mean between a and b, G:
The equation is 94.68% accurate.
Aluminum, while a dummy in the equation used to arrive at the dynamics due to asymmetry by
way of boron actually is widely used in AI because it is a conductor, which makes it an electric
shield, so it can be used to enclose electrical circuitry to protect it from electric fields. Thus we
have the two equations by molar mass and density respectively:
But
Can be written
2 ≈
Si + P
2
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
H =
2a b
a + b
G = ab
2(5.323)(5.904 − 5.7)
2.33(2.7 − 1.88) + 1.88(2.7 − 1.88)
(5.323)(2.33) = 2.2155
2.2155
2.340
= 94.68
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
4
5
∫
Ge
Si
x d x
B ≈
2Ge(Ga − As)
Si(Al − P) + P(Al − P)
GeSi
of 39 71
But
Is the geometric mean between Ge and Si. The geometric mean between a and b is given by:
Thus our equation in terms of density can be put in integral form as well:
And we see that this integral is correct:
Which is close to
94.5%
By Atomic Radius
I then considered atomic radii of these elements.
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
≈ GeSi
GeSi
G
¯
f = exp
(
1
b − a
∫
b
a
log f (x)d x
)
∫
1
0
∫
1
0
[
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
]
d x d y = exp
(
1
Ge − Si
∫
Ge
Si
log(x)d x
)
5.323(ln(5.323) − 5.323) − 2.33(ln(2.33) − 2.33) = 3.936
3.936
5.323 − 2.33
= 1.315
e
1.315
= 3.725
(5.323)(2.33) = 3.52
3.52
3.725
= 0.945
of 40 71
The atomic radii data varies some from source to source. Here I use data set 4 (previous page)
that is the average of respective values in data sets 1, 2, and 3. We have looked at molar mass,
and density and the dynamic relationships that result in terms of them do to the asymmetry
introduced into the AI elements in the periodic table. It is natural to look at radius next. It has a
lot to do with the structure and properties of the elements just like is true of the their molar
masses and densities.
Here we have the differential across Si, times Si/B plus the differential across Ge times Ge/B is
the golden ratio, phi, times the arithmetic mean between Si and Ge in atomic radius.
(Al-P)=143-116=36
of 41 71
(Ga-As)=19
(36)(115/88)=36(1.3)=46.8
19(123/88)=19(1.4)=26.6
46.8+26.6=73.4
(115+123)/2=119
The golden ratio and the golden ratio conjugate are the solution of the quadratic
that meets the conditions and a=b+c!
Where and , .
We have already said
Thus by radius the integral form of the equation is:
119
73.4
= 1.62 ≈ Φ = 1.618
(Al − P)
Si
B
+ (Ga − As)
Ge
B
= Φ
Si + Ge
2
(
a
b
)
2
−
a
b
− 1 = 0
a
b
=
b
c
Φ =
a
b
ϕ =
5 − 1
2
ϕ =
1
Φ
¯
f =
1
b − a
∫
b
a
f (x)d x
∫
1
0
∫
1
0
[
(Al − P)
Si
B
+ (Ga − As)
Ge
B
]
d x d y =
Φ
Ge − Si
∫
Ge
Si
x d x
of 42 71
The Generalized Equation
We can write a generalized equation for physical parameters whether the parameters are molar
mass, density, or atomic radius. If the differentials
and by molar mass or
and by density or
and by atomic radius
Are respectively and
And, the ratios
and by molar mass or
and by density or
and by atomic radius
Are, quotients , and , respectively, then if
And, since the geometric mean, arithmetic mean, quadratic mean (root mean square), harmonic
mean are special cases of the generalized mean:
p= 0,1, 2, -1,…
Then the generalized form of our equations is:
Where C is some constant. This is all the following equations:
(As − Ga)
(P − Al )
(Al − P)
(Al − P)
(Al − P)
(Ga − As)
ΔE
1
ΔE
2
Si
B
Ge
B
B
2Ge(Ga − As)
Si
B
2Ge(Ga − As)
P
Si
B
Ge
B
Q
1
Q
2
∇ = (ΔE
1
, ΔE
2
)
Q = (Q
1
, Q
2
)
(
1
n
n
∑
i=1
x
i
p
)
1
p
∇ ⋅ Q = C
(
1
n
n
∑
i=1
x
i
p
)
1
p
of 43 71
By molar mass.
By density.
By atomic radius.
While this is the arithmetic mean when p=1, and the harmonic mean for p=-1, I have used it as
the geometric mean for p=0, which we can see explodes to infinity at zero. But, since I am a
physicist and not a mathematician, I assume taking the limit as p—->0 is the same as evaluating
it at zero. But, as physicists undergo courses of mathematics, I do understand how to present
equations in terms of the formalities of mathematics. The proper way to treat this as a
mathematician is, the limit as p goes to zero is the geometric mean because
However we can generalize the equation to include the geometric mean without having to take
the limit as p goes to zero using the f-mean. We have:
The power mean is obtained by letting
Thus, our equation becomes:
It is the geometric mean if
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
∫
Ge
Si
dx
x
∫
1
0
∫
1
0
[
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
]
d x d y = exp
(
1
Ge − Si
∫
Ge
Si
log(x)d x
)
∫
1
0
∫
1
0
[
(Al − P)
Si
B
+ (Ga − As)
Ge
B
]
d x d y =
Φ
Ge − Si
∫
Ge
Si
x d x
M
0
(x
1
, …, x
n
) =
n
n
∏
i=1
x
i
M
f
(x
1
, …x
n
) = f
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
f (x) = x
p
∇ ⋅ Q = Cf
−1
(
1
n
⋅
n
∑
i=1
f (x
i
)
)
of 44 71
The Moon, Silicon, and Germanium
We cannot consider the equations of our three equations for molar mass and atomic radius. The
harmonic and arithmetic means respectively, and compare them to the planets, but we can
consider the equation for density. It is
Which uses the geometric mean. Thus the geometric mean between germanium and silicon is:
This is on the order of the density of the Moon which is 3.34 gram per cubic centimeter. Mars is
a little higher than our value and is 3.93 grams per cubic centimeter. The Earth is at 5.51 grams
per cubic centimeter making it much higher and, of course, planets like Jupiter and Saturn being
gas giant have very low densities (1.33 and 0.687). We have that
Where H2O is the specific gravity of water (the inverse of its density).
f (x) = log(x)
∫
1
0
∫
1
0
[
B
2Ge(Ga − As)
[
Si(Al − P) + P(Al − P)
]
]
d x d y = exp
(
1
Ge − Si
∫
Ge
Si
log(x)d x
)
(5.323)(2.33) = 3.52g /cm
3
H 2O =
B
2Ge(Ga − As)
of 45 71
The Pattern
Patterns often form with an irregularity at the first step. Like, a bulge at the center of a galaxy oar
at the center of a protoplanetary disc. Sometimes with a black hole at the center of the galaxy,
and in the case of protoplanetary discs it is where the star forms for its planetary system. Then,
have density decrease, perhaps by the exponential law:
For a disc. Or just as
It falls apart at n=-1. That is becomes
And we have to introduce the natural logarithm (log to base e) such that
We see this happens for our equations for the planetary orbits in terms of Si and Ge. They take
form at the most massive planet Jupiter, and this probably because it is the first planet after the
asteroid belt (a planet that could not form because of tidal forces on it by the other planets). That
is we have:
n=5, 6.7, 8
For the first one, i=1, then after that the pattern becomes regular and we have i=2, 4, 6,…
Thus, if we say
ρ(r) = ρ
0
e
−r /h
∫
x
n
=
x
n+1
n + 1
+ C
∫
x
n
=
x
0
0
+ C
∫
1
x
d x = ln(x) + C
P
n
=
i(Si + Ge)
2
SiG e
x =
(Si + Ge)
2
SiG e
= 4.97
of 46 71
We have
.
.
.
P
5
= i x, i = 1
P
6
= 2i x, i = 1
P
7
= 2i x, i = 2
P
8
= 2i x, i = 3
of 47 71
If you want to see the idea behind creating the orbits of the outer
planets you can do with a computer program what you can’t do with math
alone (This idea that creating output for physical reality with a few
lines of code instead of using math comes from Wolfram).
Here is my program to produce them, and you can see the logic in the
solar system as outlined in my book Planets, AI Elements, and Life, if
you know how to program in C:
// main.c
// Planets
//
// Created by Ian Beardsley on 7/30/20.
// Copyright © 2020 Ian Beardsley. All rights reserved.
//
#include <stdio.h>
int main(int argc, const char * argv[]) {
for (int i=1; i<2; i++)
{
printf("%ix\n", i);
printf("\n");
printf("\n");
for (int i=1; i<4; i++)
{
printf("%ix\n", 2*i);
}
printf("\n");
}
return 0;
}
The program outputs the following:
1x
2x
4x
6x
The following program produces the same:
// main.c
of 48 71
// The Planets
//
// Created by Ian Beardsley on 7/31/20.
// Copyright © 2020 Ian Beardsley. All rights reserved.
//
#include <stdio.h>
int main(int argc, const char * argv[]) {
//
// main.c
// Planets
//
// Created by Ian Beardsley on 7/30/20.
// Copyright © 2020 Ian Beardsley. All rights reserved.
//
int n=2;
{
for (int i=1; i<n; i++)
printf("%ix\n", i);
printf("\n");
printf("\n");
n=2*n;
for (int i=1; i<n; i++)
{
printf("%ix\n", 2*i);
}
printf("\n");
}
return 0;
}
of 49 71
Modeling The Solar System!
of 50 71
We assume the reason for the separation of the planets is due to the solution of the many body
problem, which as of yet has been solved. But the solution instinctively makes sense, a
doubling algorithm, because it kept the planets forming from the protoplanetary disc from
tearing one another apart. What I can’t explain is why it is in terms of AI elements (molar
masses)."
of 51 71
Let us hypothesize that the solar system was modeled in terms of the core AI elements, which
are silicon (Si) and germanium (Ge). Perhaps the creator was Natural and had reason for it, or
was AI. Then we start with the first planet Mercury:!
It is either the most simple combination for Si and Ge: Si+Ge, Si x Ge, Si-Ge, or Si/Ge for its
average orbital distance, or between closest and furthest approaches. It happens to be
(AU=earth-sun separation):!
(Further Approach)!
(Closer Approach)!
Now pseudocode:!
Double Both:!
, !
Take their average:!
2 !
3 !
Invert Earth:!
!
Take -2SiGe!
Change Sign:!
!
C Code:!
Int n=2;!
{!
Si
Ge
SiGe
Ge
2
+ Si
2
2Si
Ge
2SiGe
Ge
2
+ Si
2
venus =
1
Ge
2
2SiGe +
Si
3
Ge
1 +
Si
2
Ge
Ear th =
SiGe
Ge
2
− 2SiGe + Si
2
Ge
2
− 2SiGe + Si
2
SiGe
Jupiter =
Ge
2
+ 2SiGe + Si
2
SiGe
of 52 71
for (int i=1; i<n; i++)!
{!
printf (“%ix\n”, i);!
printf(“\n”);!
n=2*n;!
for (int i=1; i<n; i++)!
{!
printf (“%ix\n”,2* i);!
run:!
1x Jupiter!
2x Saturn!
4x Uranus!
6x Neptune!
!
Pseudocode:!
3 !
Move Square!
(This is the same as eliminating the central term and changing the sign of the third term)!
!
Double Quantity:!
4 !
We see the solution of the solar system is in the simplest perfect squares and quadratics:!
, , !
These are in the solutions to a differential equation of the form:!
!
By way of the method:!
x =
(Si + Ge)
2
SiGe
ear th =
SiGe
(Ge − Si )
2
SiGe
Ge
2
− Si
2
m ars =
2SiGe
Ge
2
− Si
2
(Ge − Si )
2
(Ge + Si )
2
(Ge − Si )(Ge + Si)
d
2
y
dx
2
− 2
dy
dx
+ y = 0
of 53 71
!
!
!
!
!
!
This is essentially the Titius-Bode Rule for the distribution of the planets which is:!
!
Where Mercury is , Venus is n=0, Earth is n=1, and so on where we include the
asteroid belt and r is in 1 E11 m. It starts to fall apart for Uranus and Neptune, but this I believe
is because we need to break our equation into two equations on either side of the asteroid belt
where the planet didn’t form (broke up into asteroids).!
In this sense we see this may be the solution to the many body problem for our solar system
by looking at it as a thin protoplanetary disc composed of silicon and germanium. This is for
our quadratic!
!
For our quadratic!
!
We have!
!
!
Which is the exponential law for density distribution in a disk:!
!
y = e
mx
y′ = m e
mx
y′ ′ = m
2
e
mx
(m − 1)(m − 1) = 0
y = c
1
e
x
y = c
2
xe
x
r = 0.6 + 0.45(2
n
)
n = − ∞
(Ge − Si )
2
= (Ge − Si)(Ge − Si)
(Ge + Si )
2
= (Ge + Si)(Ge + Si )
y = c
1
e
−x
y = c
2
xe
−x
ρ(r) = ρ
0
e
−r /h
of 54 71
Hydrogen"
of 55 71
I suggested that in Nature, the first step is often outside the characteristics of the ensuing
pattern. For example, the bulge at the center of the galaxy in density and size, from which after
follows a smoother decrease in density, or the bulge at the center of a protoplanetary cloud,
like that which would seem to exist in the protoplanetary disc around a companion star in
epsilon aurigae.!
Or, as in mathematics where
It falls apart at n=-1. That is becomes
And we have to introduce the natural logarithm (log to base e) such that
We can look at this in the biological as well. Consider how a plant often has a flower at the top
of the stem, but the pattern begins with the leaves spiraling down the stem with periodicity of
the golden ratio (phi) at right angles to it,!
This can be understood through the science of computers. In order to make a loop, we need to
iterate a series of steps until a condition is met. For example in C:!
While (i<5)!
{!
i=i+1!
{!
Printf(“*”);!
}!
It outputs five stars:!
* * * * *!
Notice the condition that must be met before the loop terminates is While i less than 5 (i<5). Yet
it counts to five inclusive. This is because the C computer language is zero indexed, which
means it starts counting from zero as the first step unless told otherwise, so less than five it
counts to five. If C was indexed to start counting at one I would write the program:!
While (i<6)!
{!
i=i+1!
{!
Printf(“ * ”;!
}!
∫
x
n
=
x
n+1
n + 1
+ C
∫
x
n
=
x
0
0
+ C
∫
1
x
d x = ln(x) + C
of 56 71
Thus I would like to suggest the terrestrial planets interior to the asteroid belt are like the flower,
and the outer planets, beginning with Jupiter are the leaves where the pattern begins.!
As such, since our polynomials in terms of the AI elements are the simplest of perfect squares
and quadratics:!
(Si+Ge)(Si+Ge), (Si-Ge)(Si-Ge), (Si-Ge)(Si+Ge)!
In that Si and Ge have coefficients of one, then let us consider the solar system in terms of the
simplest atom, hydrogen H, which is one proton and one electron, and in that it is the most
abundant element in the Universe and certainly was central to the protoplanetary disc. We look
at the atomic spectra of hydrogen:!
And pull out the Balmer Series, because it is in visible range to the human eye:!
!
The wavelengths of its emmision spectra are given by:!
!
Rydberg constant!
1
λ
= R
H
(
1
n
1
2
−
1
n
1
2
)
R
H
=
of 57 71
!
There are four visible lines in the Balmer series (n>2; n=2) and they are!
410nm!
434nm!
486nm!
656nm!
These correspond to!
!
E=(6.625E-34)(410E-9m)=!
2.7-67E-40 Joules…!
2.8757E-40 Joules!
3.22E-40 Joules!
4.3467E-40 Joules!
But if E is potential in an inverse square field%
%
!
!
charge of proton!
Then we have!
!
Produces,…!
!
…!
r=8.00E11m!
r=7.147E11m!
r=5.2947E11m!
Take the geometric mean between the two central terms!
Jupiter Orbit!
λ =
c
ν
E = h ν
F = k
e
q
1
q
2
r
2
k
e
=
1
4πe
0
= 9E 9
q
1
= q
2
= 1.6E − 19 =
r = k
q
1
q
2
E
r =
(8.99E 9)(1.6E − 19)
2
2.87575E − 40J
= 8.47E11m
(7.1)(8.00) = 7.5E11m ≈
of 58 71
The average orbital distance of Jupiter from the sun is 7.78E11m.!
I find this very interesting as Jupiter carries most of the mass of the solar system not counting
the sun, and thus most of its evolutionary dynamics."
of 59 71
The Scale of Things!
of 60 71
A slowly rotating cloud of gas and dust around a protostar collapses into a disc flattening out
and the protostar warms up and eventually ignites going into its phase of burning hydrogen to
make helium. It blows out the lighter elements close in and planets begin to form. The heavier
terrestrial planets are on the inside, the gas giants on the outside.!
The star has a halo formed around it of gas and dust that can be likened to a fairly thin, flat,
right circular cylinder with a large hole in the center. The gases are mostly hydrogen and helium
and can be thought to begin where Jupiter is and goes out to the edge where Pluto is. The
majority of the mass of the hydrogen and helium is at what is to become Jupiter and Saturn
orbits because they carry the majority of the mass of the solar system significantly.!
Thus, from this, we can get an idea of the order of density we are speaking of, of just the
hydrogen aspect of the halo, by balancing the pressure due to temperature at Jupiter orbit, and
the pressure due to the Sun’s gravity at Jupiter as well, and considering the volume of a disk
one astronomical unit thick form Jupiter orbit to Saturn orbit.!
The luminosity of the sun is now, but was warmer in its infancy:!
!
The solar constant at Jupiter orbit is:!
!
The temperature is derived from the steffan-boltzman law for blackbody radiators:!
!
Where sigma is the Stefan Boltzmann constant:!
!
Thus,…!
L
0
= 3.9 × 10
26
J/s
S
0
=
3.9 × 10
26
4π(7.78 × 10
11
)
2
= 51.27
wat ts
m
2
σ T
4
= S
0
σ = 5.67 × 10
−8
of 61 71
!
This is,…!
!
Jupiter is 89% hydrogen (H2) gas and 10% Helium (He) gas. The mass of Jupiter is 1.8982E27
kg. This is in terms of hydrogen:!
!
Saturn is 96.3% Hydrogen (H2) gas. The mass of Saturn is 5.683E26 kg. This is in terms of
hydrogen:!
0.963(5.6834E26kg)=5.473E26kg!
The mass of H2 in these taken together is:!
1.689E27kg+5.473E26kg=2.2363E27kg.!
The volume of the halo staring from Jupiter (R1) and ending at Saturn (R2) for a thickness of
1AU, the earth-sun separation (1.5E11m)…!
!
The pressure of a gas on earth given altitude is related by%
%
!
If g is the gravity at Jupiter due to the Sun and rho the density of the hydrogen halo which is
given by:!
!
Then from!
!
We have!
!
Thus,…!
T =
(
51.27
5.67 × 10
−8
)
1
4
= 173.41
∘
K
T = 173.41 − 273 = − 100
∘
C
0.89(1.8982E 27kg) = 1.689E 27kg
V = πh(R
2
2
− R
2
1
) = 3(1.5E11)
[
2E 24 − 6E 23
]
= 6.3E 35m
3
dP
dh
= ρg
ρ =
2.2363E 26kg
6.3E35m
3
= 3.55E − 9kg/m
3
g =
M
r
2
G
g =
2E 30kg
(7.78E11)
2
6.674E − 11 = 0.00022
m
s
2
of 62 71
~0.55Pa!
The Pressure at Jupiter due to temperature determines the density of the protoplanetary disc
by the ideal gas law!
!
Where!
!
!
Or,…!
!
H2=2(1.01g/mol)!
!
To get an idea of what kind of density this is compare it to the density of H2 at 0 degrees
centigrade and 1 atmosphere of pressure (standard temperature and pressure, STP):!
0.08988g/L!
We can convert our 0.55 Pascals to bar !
!
0.55Pa=5.5e-6bar=0.0000055bar !
Thus,…!
!
Where a bar is about the pressure at see level and is 1,000,000 dynes per square centimeter.
Thus,…!
!
Realistically, the pressure is different at right angles to the disc than out along its plane radially
in all directions, but this gives one an idea of scale of thing (or order of magnitude)."
P = ρgr = (3.55E − 9)(0.00022)(7.78E11) = 0.5467Pa
PV = n RT
R = 8.314m
3
⋅ Pa ⋅ K
−1
⋅ mol
−1
V
n
=
(8.314)(173
∘K
)
(0.55Pa)
= 2615.13L /m ol
n
V
= 0.00038moles /L
(
0.00038
mol
L
)(
2.02
g
mol
)
= 0.0007676
g
L
H
2
1Pa = 1e − 5bar
P = 5.5μbar
(0.0000055bar)(1,000,000d yn)
cm
2
= 5.5
dyn
cm
2
of 63 71
The Nature of our Equation!
of 64 71
When we say that!
!
!
Can be equated to solving!
and !
With m=-1 and m=1 and that this is the solution to the differential equation of the form:!
!
Which is!
; ; ; !
Which is the solution to the Titius-Bode rule!
!
And, the exponential law for density distribution in the protoplanetary disc!
How do we see this?
We are used to solving
By completing the square or factoring to write it as
and we have
However, if we want to solve
By solving
Since in the latter x=-3, in the former it factors into
(Ge + Si )
2
= Ge
2
+ 2SiGe + Si
2
(Ge − Si )
2
= Si
2
− 2SiGe + Ge
2
(m + 1)
2
(m − 1)
2
d
2
y
dx
2
− 2
dy
dx
+ y = 0
y = c
1
e
x
y = c
2
xe
x
y = c
1
e
−x
y = c
2
xe
−x
r = 0.6 + 0.45(2
n
)
ρ(r) = ρ
0
e
−r /h
x
2
+ 6x + 9 = 0
(x + 3) = 0
x = − 3
x
2
+ 3x y + 9y
2
= 0
x
2
+ 6x + 9 = 0
of 65 71
We can solve it by by merely solving
And the 3 is the coefficient of y in .
We want to solve
Which is:
If Si is y and Ge is x, then
Thus we have produced our
In our Mars equation
And we already had our .
And in the Jupiter through Neptune equations
We find that if in our quadratic
Has solution
(x + 3y)(x + 3y)
(x + 3)(x + 3) = 0
(x + 3y)(x + 3y)
x
2
+ 2x y + y
2
= 0
−y
2
+ x
2
(x + y)
=
±
1
Ge
2
− Si
2
(Ge + Si )
=
±
1
Ge
2
− Si
2
1.52 =
2SiGe
Ge
2
− Si
2
(Ge + Si )
2
P
n
=
i(Si + Ge)
2
SiG e
Ge
2
+ 2SiGe + Ge
2
= 0
of 66 71
!
Which actually works for Si/Ge in both molar mass and density (The density of Si is 2.33 grams
per cubic centimeter and Ge is 5.323 grams per cubic centimeter).!
The interesting thing is that as we showed earlier, if HA (hydroxyapeatite) is the mineral
component of bone, Then!
!
!
Which is the sextic!
!
Which has the solution!
!
As well as x=y=-1, x=y=-1!
The values of the m in !
!
!
For the solutions of the differential equations of the form!
That produce our protoplanetary solution.
But, it goes further. Hydroxyapatite is
!
However, the organic component of bone is!
!
Si
Ge
≈
1
2 + 1
Si
HA
Si +
[
1 −
Si
HA
]
Ge = H A
2SiGe
Si + Ge
≈ H A
x
2
(x + y)
4
− x y(x + y)
4
+ 2x y
2
(x + y)
3
− 4x
2
y
2
(x + y)
2
= 0
Si
Ge
≈
1
2 + 1
(m − 1)(m − 1) = 0
(m + 1)(m + 1) = 0
d
2
y
dx
2
− 2
dy
dx
+ y = 0
Ca
5
(PO
4
)
3
OH = 502.32
g
mol
C
57
H
91
N
19
O
16
= 1298.67
g
mol
of 67 71
We have!
!
The golden ratio conjugate is!
!
But,…!
!
However,…!
!
So,…!
!
So the basic framework for the solar system is described by the basic framework for
vertebrates, the skeleton made from bone.!
This is all intrinsically in the binomial theorem which is:!
!
Where we have considered n=2. However for the purposes n other than 2 it is useful because
we can refine the accuracy of our polynomials using values other than 2."
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= 0.386795722
ϕ = 0.618033989
1 − ϕ = 0.381966011
Si
Ge
=
28.09
72.61
= 0.386861314 ≈ (1 − ϕ)
Si
Ge
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
(x + y)
2
= x
r
+ rx
r−1
y +
r(r − 1)
2!
+
r(r − 1)(r − 2)
3!
x
r−3
y
3
+ …
of 68 71
The Orbits of the Planets!
of 69 71
We have said the equations for the evolution of the Solar System are of the form!
; !
And !
!
We here note that!
!
The decimal part of the square root of 2. We have also said!
!
!
!
!
This means:!
!
This results in the following simple equations for the planets:!
!
!
!
!
!
!
y = Ce
x
y = Cx
−x
Si
Ge
≈
1
2 + 1
1
2 + 1
= 0.414
(1 − ϕ) = 0.38
Si
Ge
=
28.09
72.61
= 0.387
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= 0.387
Si
Ge
≈
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
e
Si
Ge
= e
Ca
5
(PO
4
)
3
OH
C
57
H
91
N
19
O
16
= e
(1−ϕ)
(1 − ϕ)e
ϕ
= 0.7AU = Venus
ϕe
(1−ϕ)
= 0.9AU = Ear th
ϕ
2
e
(2−ϕ)
= 1.52 = Mars
2ϕe
(2−ϕ)
= 4.9 = Jupiter
4ϕe
(2−ϕ)
= 10 = Sat ur n
6ϕe
(2−ϕ)
= 14.768 = Uranus
8ϕe
(2−ϕ)
= 19.69 = Neptune
of 70 71
Proposal For Mars Mission Logo
Here I propose a logo for a Mars mission. The forms as symbols for the planets Earth and
Mars in the logo originate from a theory of my own where I put forward mathematical
expressions for the planets that are related to one another in two equations, one for the
inner planets, the other for the outer planets. The expressions for the planets are all
described in terms of the two primary semiconductor elements - silicon and germanium -
which are primary to make artificial intelligence (AI). These AI elements are described in
the theory in their relationship to biological life. The equations generate three dimensional
forms for each planet, that embody their place in the cosmic scheme and express the
connection, and indeed the idea, that we have a place and purpose in this solar system. I
therefore feel they are perfect for use in the design of a logo for a Mars Mission. Here, I
present the concept for a logo, using the forms in my theory for Earth and Mars, and a
symbolic image of a rocket. This is the basic concept for the logo, the theory from which
it is derived is was presented in this paper: "
of 71 71
The Author!